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Sloan [31]
2 years ago
12

A quantity that has only magnitude(size) is

Physics
1 answer:
KatRina [158]2 years ago
8 0
Scalar quantities- I think that’s the que ur asking
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Write any two difference between CGS and MKS system of measurement ?​
Tresset [83]

Answer:

MKS stands for Meter, Kilogram and second. In this system of unit mass is given in Kilogram, length in meter and time in second. ... CGS system stands for Centimeter- Gram- Second system. In CGS system, length is measured in centimeters mass is measured in grams and time is in seconds.

7 0
3 years ago
If the magnetic flux through a loop increases according to the relation; where , is in milliweber (mWb) and t is in seconds. Cal
Alekssandra [29.7K]

The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

<h3>emf induced in the loop</h3>

The magnitude of e.m.f induced in the loop is calculated as follows;

emf = dФ/dt

Ф = 6t² + 7t

dФ/dt = 12t + 7

at t = 2 seconds

emf = dФ/dt = 12(2) + 7 = 31 V

Thus,  the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

Learn more about emf here: brainly.com/question/24158806

#SPJ1

6 0
2 years ago
The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on
choli [55]

Answer:

25 times the average speed

7 1
3 years ago
Read 2 more answers
A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.
snow_lady [41]

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is  v =  3.79 *10^{5} \ m/s  

Explanation:

From the question we are told that

    The  magnitude of the electric field is  E  =  144 \ kV /m   =  144*10^{3} \  V/m

     The magnetic field is  B  =  0.38 \ T

   

The force due to the electric field is mathematically represented as

      F_e =  E  * q

and

The force due to the magnetic field is mathematically represented as

    F_b  =  q * v  *  B * sin(\theta )

Now given that it is perpendicular ,  \theta  = 90

=>   F_b  =  q * v  *  B * sin(90)

=>   F_b  =  q * v  *  B

Now  given that it is not deflected it means that

        F_ e  =  F_b

=>    q *  E = q *  v  * B

=>   v =  \frac{E}{B }

 substituting values

     v =  \frac{ 144 *10^{3}}{0.38 }

     v =  3.79 *10^{5} \ m/s

7 0
3 years ago
a student attaches a 0.5 kg object to a 0.7 m string and rotates the object around her head and parallel to the ground. how much
Marina CMI [18]

The object would have a centripetal acceleration <em>a</em> of

<em>a</em> = (12 m/s)² / (0.7 m) ≈ 205.714 m/s²

so that the required tension in the string would be

<em>T</em> = (0.5 kg) <em>a</em> ≈ 102.857 N ≈ 100 N

(rounding to 1 significant digit)

3 0
3 years ago
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